Question

In a photoelectric experiment it is found that a stopping potential of 1.00 V is needed to stop all the electrons when incident light of wavelength 278 nm is used and 2.6 V is needed for light of wavelength 207 nm. From these data determine Planck's constant and the work function of the metal.

Answer 1

Planck’s constant is
`6.9*10^(-34) Js`

work function of metal is
`5.846*10^(-19)`

Step-by-step explanation:

The maximum kinetic energy from electron from photoelectric effect is given by

`\K_(max)=eV_o` where
`V_o` is applied voltage and e is charge on electron and substituting charge on electron by
`1.6*10^(-19)` and 1.00 V for
`V_o`

`K_(max)=1.6*10^(-19)*1=1.6*10^(-19) J`

Considering that the wavelength, \lambda of light used is given as
`278*10^(-9) m`

Energy of light is given by

`E=\frac {hc}{ \lambda}` where
`\lambda` is the wavelength, h is Planck’s constant and c is speed of light. Taking c for
`3*10^(8) m/s`

Substituting values of wavelength and speed of light we obtain

`E=\frac {h*3*10^(8)}{278*10^(-9)}=h1.07914*10^(15) J`

If work function is
`\phi` then

`E=\phi + k_(max)` hence

`h1.07914*10^(15) J=\phi +1.6*10^(-19) J` ------- Equation 1

Energy corresponding to wavelength of 207 nm is

`E=h\frac {3*10^(8) m/s}{207*10^(-9)}=h(1.45*10^(15)}) J`

Maximum kinetic energy of electrons when
`V_o` is 2.6 V becomes

`K_(max)=(1.6*10^(-19))*2.6V=4.16*10^(-19) J`

From
`E=\phi + k_(max)` and substituting
`h(1.45*10^(15)}) J` for E and
`4.16*10^(-19) J` for
`K_(max)` we have

`h(1.45*10^(15)}) J=\phi +4.16*10^(-19) J` ------ Equation 2

Equation 2 minus equation 1 gives

`h3.71*10^(14)=2.56*10^(-19)`

`h=\frac {2.56*10^(-19)}{3.71*10^(14)}\approx 6.9*10^(-34) Js`

Therefore, Planck’s constant is
`6.9*10^(-34) Js`

Recalling equation 1 and substituting back the value of h as obtained

`h1.07914*10^(15) J=\phi +1.6*10^(-19) J`

`(6.9*10^(-34))*(1.07914*10^(15))=\phi +1.6*10^(-19) J`

`\ph=(6.9*10^(-34))*(1.07914*10^(15))-(1.6*10^(-19))=5.846*10^(-19)`

Therefore, work function of metal is
`5.846*10^(-19)`